Is $X Y/|Y|$, where $X, Y \text{ i.i.d. } \sim N(0, 1)$, normally distributed?

Question

In this answer, it was shown that the product of two i.i.d. standard normal r.v.s is not normally distributed. What if we scale one of them? In other words, if $X, Y \text{ i.i.d } \sim N(0, 1)$, what is the distribution of $Z := X\frac{Y}{|Y|}$? By looking at he distribution function of $Z$ (though I didn't finish the calculation), $Z$ is unlikely to be normal. However, the simulation seems to suggest that $Z$ is normal (the QQ plot of $10,000$ realizations of $Z$ is shown below). Is there a clever way to prove or disprove $Z$ is normally distributed?

Answer 1

Conditional on $Y = y$, $Z$ is $\mathcal N(0,1)$. Since the regular conditional distribution of $Z$ is the same for all $y$, we conclude $Z$ is independent of $Y$. In particular, $Z\sim \mathcal N(0,1)$.

Generalizing, if $X\sim \mathcal N(0,\sigma^2_1),Y\sim \mathcal N(0,\sigma^2_2)$ one still has $Z\sim \mathcal N(0,\sigma_1^2)$. It is easy to see that this fails if $E(Y)\neq 0$.

Answer 2

As an alternative method to the accepted answer, the proof can be completed by the following two steps:

1. $U := Y/|Y|$ takes values $\pm 1$ almost surely, and $P(U = -1) = P(U = 1) = \frac{1}{2}$.
2. If $P(U = -1) = P(U = 1) = \frac{1}{2}$ and $X \sim N(0, 1)$ are independent, then $UX \sim N(0, 1)$.

Note that property $1$ holds for any r.v. $Y$ such that $P[Y < 0] = P[Y > 0] = 1/2$ (this condition implies that $Y$ has no mass at $0$: as $P[Y = 0] = 1 - P[Y > 0] - P[Y < 0] = 0$ -- it can be seen that the $Y \sim N(0, 1)$ in the OP trivially satisfies this condition). Indeed, \begin{align*} & P[U = 1] = P[Y > 0] = \frac{1}{2}, \\ & P[U = -1] = P[Y < 0] = \frac{1}{2}. \end{align*}

Property 2 holds because for any $x \in \mathbb{R}$: \begin{align*} P[UX \leq x ] &= P[UX \leq x, U = 1] + P[UX \leq x, U = -1] \\ &= P[X \leq x, U = 1] + P[X \geq -x, U = -1] \\ &= P[X \leq x]P[U = 1] + P[X \geq -x]P[U = -1] \tag{by independence} \\ &= \Phi(x) \times \frac{1}{2} + (1 - \Phi(-x)) \times \frac{1}{2} \\ &= \Phi(x). \end{align*}